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Iron oxide ores are reduced to iron metal by exothermic reaction with carbon monoxide. $$ \mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) $$ Which of the following changes in condition will cause the equilibrium to shift to the right? a. Add \(\mathrm{CO}_{2}\) b. Add \(\mathrm{FeO}\) c. Add CO d. Raise the temperature

Short Answer

Expert verified
Adding FeO or CO will shift the equilibrium to the right.

Step by step solution

01

Identify the Direction of Equilibrium Shift

The equilibrium of a reaction will shift to either the right (products) or left (reactants) to counteract any change applied to the system according to Le Chatelier's principle. In this case, we are interested in changes that will shift the equilibrium to the right, producing more iron.
02

Analyze the Effect of Adding COâ‚‚

Adding \( ext{CO}_{2}\) will increase the concentration of products in the reaction. According to Le Chatelier's principle, the equilibrium will shift to the left to decrease the concentration of products.
03

Analyze the Effect of Adding FeO

Adding \( ext{FeO}\) will increase the concentration of reactants. According to Le Chatelier's principle, the equilibrium will shift to the right to consume the added reactant, producing more products.
04

Analyze the Effect of Adding CO

Adding CO increases the concentration of reactants, prompting the equilibrium to shift to the right in order to reduce the added reactant and produce more \( ext{Fe}\) and \( ext{CO}_{2}\).
05

Analyze the Effect of Raising Temperature

The reaction is exothermic, meaning heat is released by the reaction. Increasing the temperature will add heat, causing the equilibrium to shift to the left to absorb the excess heat and counteract the added energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle describes how a chemical system at equilibrium responds to changes in concentration, pressure, or temperature. It provides a predictive framework for determining the direction in which the equilibrium will shift to counteract the imposed change. When a system at equilibrium experiences a change, the system will adjust to minimize that change.
In the context of the exercise, we explore several scenarios to identify which change will push the equilibrium towards producing more iron. Understanding how equilibrium reacts to these changes is essential for efficiently regulating and predicting chemical processes.
Exothermic Reaction
An exothermic reaction is one that releases heat into the surroundings. In the iron reduction process involving carbon monoxide and iron oxide, the reaction is exothermic. This means that as iron and carbon dioxide are produced, heat is also released.
When considering how temperature affects this reaction, it's crucial to remember that adding heat to an exothermic reaction will shift the equilibrium towards the reactants. Raising the temperature will increase the system's heat content and cause the reaction to move in the direction that absorbs heat, which is to revert to the reactants, reducing further iron formation.
Iron Reduction
The reduction of iron from its ore is an essential process in metallurgy. It involves a chemical reaction where iron oxide is converted into iron metal, facilitated by carbon monoxide. This conversion is crucial for iron extraction since it transforms the oxide form of iron in the ore into elemental iron that can be used in various applications.
In the exercise, the reaction is symbolized as: \[ \mathrm{FeO} (\mathrm{s}) + \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) \] Understanding this process helps in managing and improving the efficiency of industrial iron production, which hinges significantly on the right conditions favoring the desired reaction.
Reaction Shift Analysis
Reaction shift analysis involves understanding how changes in reactants or products influence the equilibrium of a chemical reaction. In Le Chatelier's Principle, adding substances or altering conditions will induce the equilibrium to shift either to the left (favoring reactants) or right (favoring products) as the system attempts to counteract the adjustment.
For the exercise:
  • Adding \(\mathrm{FeO}\) or \(\mathrm{CO}\) will cause a shift to the right, enhancing production of \(\mathrm{Fe}\) and \(\mathrm{CO}_2\).
  • Adding \(\mathrm{CO}_2\) shifts to the left, decreasing \(\mathrm{Fe}\) production.
  • Raising the temperature also shifts left due to the exothermic nature of the reaction.
Each scenario involves a precise analysis of how equilibrium adjustments respond to changes in reactant and product concentrations, thereby emphasizing the dynamic balance within reversible reactions.

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Most popular questions from this chapter

Match the following: Column I Column II A. \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}\) (p) \(\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}\) (g) B. \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons\) (q) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}\) \(2 \mathrm{SO}_{3}(\mathrm{~g})\) C. \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\) (r) \(\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}}\) \(\mathrm{Cl}_{2}(\mathrm{~g})\) D. \(2 \mathrm{KClO}_{3} 2 \mathrm{KCl}+3 \mathrm{O}_{2}\) (s) High temperature (t) Irreversible reaction

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) undergoes a molecular rearrangement in the presence of \(\mathrm{AlCl}_{3}\) to form methylcyclopentane \(\left(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right)\), according to the equation \(\mathrm{C}_{6} \mathrm{H}_{12} \rightleftharpoons \mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) If \(\mathrm{K}_{\mathrm{C}}=0.143\) at \(25^{\circ} \mathrm{C}\) for this reaction, find the equilibrium concentrations of \(\mathrm{C}_{6} \mathrm{H}_{12}\) and \(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) if the initial concentrations are \(0.200 \mathrm{M}\) and \(0.100\) M respectively a. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.286\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.016 \mathrm{M}\) b. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.262\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.038 \mathrm{M}\) c. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.186\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.162 \mathrm{M}\) d. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.164\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.621 \mathrm{M}\)

Match the following: Column I \(\quad\) Column II (p) Increase of pressure A. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\) favours forward reaction B. \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow\) (q) Constant pressure \(2 \mathrm{NO}(\mathrm{g})\) favours forward reactior C. \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{~s}) \rightleftharpoons\) (r) D. \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s}) \quad(\mathrm{s})\) \(+\mathrm{CO}_{2}(\mathrm{~g})\)

For the reaction $$ \mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ the forward reaction at constant temperature is favoured by a. Increasing the volume of the container b. Introducing \(\mathrm{PCl}_{5}\) at constant volume c. Introducing an inert gas at constant volume d. Introducing \(\mathrm{Cl}_{2}(\mathrm{~g})\) at constant volume

Match the following: Column I Column II A. Law of mass action (p) Guldberg wage B. Active mass of \(\mathrm{CaCO}_{3}\) (q) \(\mathrm{Kp}=\mathrm{Kc}(\mathrm{RT})\) (s) C. \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{SO}_{3}\) (r) 1 D. \(\mathrm{PCl}_{5} \leftrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (s) zero (t) \(\mathrm{Kp}=\mathrm{Kc}(\mathrm{RT})^{-1}\)

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